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poizon [28]
4 years ago
13

Which of the following has the most biomass? a. quaternary consumers b. secondary consumers c. primary consumers d. producers

Physics
2 answers:
elena-s [515]4 years ago
4 0
<h3><u>Answer;</u></h3>

<em>Producers </em>

<h3><u>Explanation;</u></h3>
  • <u><em>Producers </em></u>occupy the lowest level in any food chain or food web. They are the organisms that all other organisms in the food chain rely on.
  • <em><u>Producers </u></em>have the ability to make their own food through the process of photosynthesis where they use energy from sunlight, together with water and carbon dioxide to generate food.
  • The ability of producers to make their own food and the fact that they occupy the lowest level in a food chain means <em><u>they have the highest biomass.</u></em>
elena55 [62]4 years ago
4 0

Answer:

Producers

Explanation:

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If an object weighs 300 N on earth, what is it’s mass on the moon?
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The mass of the object on the Moon (and anywhere else) is about 30.61kg. Please see more detail below.

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Weight is the gravitational force exerted on the object and is a function of mass and gravitational acceleration:

(weight) = (mass) x (gravitational acceleration)

We are to find the mass, knowing the weight on Earth to be 300N:

(mass) = (weight on Earth) / (gravitational acceleration on Earth) = 300N / 9.8 m/s^2 = 30.61 kg

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The mass of the object is independent of gravity. Therefore the answer to the question "What is its mass on the Moon" is 30.61kg.

If the question were what is its weight on the Moon, the answer would be

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(a) 1.49 m/s

The conservation of momentum states that the total initial momentum is equal to the total final momentum:

p_i = p_f\\m u_b + M u_B = (m+M)v

where

m = 0.016 kg is the mass of the bullet

u_b = 280 m/s is the initial velocity of the bullet

M = 3 kg is the mass of the block

u_B = 0 is the initial velocity of the block

v = ? is the final velocity of the block and the bullet

Solving the equation for v, we find

v=\frac{m u_b}{m+M}=\frac{(0.016 kg)(280 m/s)}{0.016 kg+3 kg}=1.49 m/s

(b) Before: 627.2 J, after: 3.3 J

The initial kinetic energy is (it is just the one of the bullet, since the block is at rest):

K_i = \frac{1}{2}mu_b^2 = \frac{1}{2}(0.016 kg)(280 m/s)^2=627.2 J

The final kinetic energy is the kinetic energy of the bullet+block system after the collision:

K_f = \frac{1}{2}(m+M)v^2=\frac{1}{2}(0.016 kg+3 kg)(1.49 m/s)^2=3.3 J

(c) The Energy Principle isn't valid for an inelastic collision.

In fact, during an inelastic collision, the total momentum of the system is conserved, while the total kinetic energy is not: this means that part of the kinetic energy of the system is losted in the collision. The principle of conservation of energy, however, is still valid: in fact, the energy has not been simply lost, but it has been converted into other forms of energy (thermal energy).

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4 years ago
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